Question

If we have an algorithm inequality ${{\log }_{\dfrac{1}{\sqrt{2}}}}\sin x>0$, $x\in \left[ 0,4\pi \right]$, then find the number of values of x which are integral multiples of $\dfrac{\pi }{4}$?
(a) 4
(b) 12
(c) 3
(d) None of these.

Answer 1

We start solving the by finding the interval for $\sin x$ using the given condition ${{\log }_{\dfrac{1}{\sqrt{2}}}}\sin x>0$. We also find another interval for $\sin x$ using the that in a function ${{\log }_{a}}x$ for any value $a>0$, $x>0$. Using these two intervals for $\sin x$, we find the interval for ‘x’. From the interval of ‘x’, we find the multiples of $\dfrac{\pi }{4}$ and find the number of multiples present.

Complete step-by-step solution:
Given that we have ${{\log }_{\dfrac{1}{\sqrt{2}}}}\sin x>0$, and the values of x lies in the interval $\left[ 0,4\pi \right]$. We need to find the number of values of x which are integral multiples of $\dfrac{\pi }{4}$.
Let us start by solving for solution set of ‘x’, if the given condition ${{\log }_{\dfrac{1}{\sqrt{2}}}}\sin x>0$ holds true.
So, we have got ${{\log }_{\dfrac{1}{\sqrt{2}}}}\sin x>0$ ---(1).
We know that ${{\log }_{\dfrac{1}{a}}}x=-{{\log }_{a}}x$. We use this result in equation (1).
So, we have got $-{{\log }_{\sqrt{2}}}\sin x>0$.
We have got ${{\log }_{\sqrt{2}}}\sin x<0$.
We know that ${{\log }_{{{a}^{n}}}}x=\dfrac{1}{n}\times {{\log }_{a}}x$.
We have got ${{\log }_{{{2}^{\dfrac{1}{2}}}}}\sin x<0$.
We have got $\dfrac{1}{\dfrac{1}{2}}\times {{\log }_{2}}\sin x<0$.
We have got $2\times {{\log }_{2}}\sin x<0$.
We have got ${{\log }_{2}}\sin x<0$.
We know that if ${{\log }_{a}}b=c$, then ${{a}^{c}}=b$.
We have got $\sin x<{{2}^{0}}$.
We have got $\sin x<1$ ---(1).
We also know that in a function ${{\log }_{a}}x$ for any value $a>0$, x should be greater than 0 $\left( x>0 \right)$.
So, we have got $\sin x>0$ ---(2).
From equations (1) and (2), we got the values of $\sin x$ lies in the interval $\left( 0,1 \right)$. We know that $\sin x>0$ in the interval $x\in \left( 0,\pi \right)\cup \left( 2\pi ,3\pi \right)$. But the value of $\sin x$ at $x=\dfrac{\pi }{2}$ and $x=\dfrac{5\pi }{2}$ is ‘1’.
So, we got the interval for x as follows $x\in \left( 0,\dfrac{\pi }{2} \right)\cup \left( \dfrac{\pi }{2},\pi \right)\cup \left( 2\pi ,\dfrac{5\pi }{2} \right)\cup \left( \dfrac{5\pi }{2},3\pi \right)$.
The multiples of $\dfrac{\pi }{4}$ from the interval $\left( 0,\dfrac{\pi }{2} \right)\cup \left( \dfrac{\pi }{2},\pi \right)\cup \left( 2\pi ,\dfrac{5\pi }{2} \right)\cup \left( \dfrac{5\pi }{2},3\pi \right)$ are $\dfrac{\pi }{4},\dfrac{3\pi }{4},\dfrac{9\pi }{4},\dfrac{13\pi }{4}$.
We got a total of ‘4’ multiples of $\dfrac{\pi }{4}$.
$\therefore$ The number of values of x which are integral multiples of $\dfrac{\pi }{4}$ is 4.
The correct for the given problem is (a).

Note: We should forget to take the condition for logarithmic function, as it gives us the main condition required for the problem. Since the value of x lies in the interval $x\in \left[ 0,4\pi \right]$, we have got 4 solutions. If interval changes, the no. of solutions differs.